Question

Sin (150)- tan (315) +cos (300) +sec ^2 (360)

Answer 1

Sin (150)- tan (315) +cos (300) + sec ^2 (360) = 2

Step-by-step explanation:

Sin (150) = Sin (180 -30⁰) = sin 30⁰ =1/2

tan (315) = tan (360-45⁰)= - tan 45⁰ = -1

cos (300) = cos (360- 60⁰) = cos60⁰ =1/2

sec (360) = 0

sec ^2 (360) =0

now,

Sin (150)- tan (315) +cos (300) +sec ^2 (360)

$\frac{1}{2} +1 + \frac{1}{2} + 0$

= 1+1

=2

Answer 2

Answer:

Required value of the given term is 2.

Step-by-step explanation:

Given,

$\sin( {150}^{o} ) - \tan( {315}^{o} ) + \cos( {300}^{o} ) {sec}^{2} ( {360}^{o} )$

Now,

$\sin( {150}^{o} ) \\ = \sin( {180}^{o} - {30}^{o} ) \\ = \sin( {30}^{o} ) \\ = \frac{1}{2}$

and

$\tan( {315}^{o} ) \\ = \tan( {360}^{o} - {45}^{o} ) \\ = -\tan( {45}^{o}) \\ = - 1$

and

$\cos( {300}^{o} ) \\ = \cos( {360}^{o} - {60}^{o} ) \\ = \cos( {60}^{o}) \\ = \frac{1}{2}$

and

$\sec( {360}^{o} ) = 0 \\ {sec}^{2} {360}^{o} = 0$

Now,

$\sin( {150}^{o} ) - \tan( {315}^{o} ) + \cos( {300}^{o} ) {sec}^{2} ( {360}^{o} ) \\ = \frac{1}{2} + 1 + \frac{1}{2} + 0 \\ = 1 + 1 \\ = 2$

This is a problem of Trigonometry.

Some important formulas of Trigonometry formula:

$sin(x) = \cos(\frac{\pi}{2} - x) \\ \tan(x) = \cot(\frac{\pi}{2} - x) \\ \sec(x) = \csc(\frac{\pi}{2} - x) \\ \cos(x) = \sin(\frac{\pi}{2} - x) \\ \cot(x) = \tan(\frac{\pi}{2} - x) \\ \csc(x) = \sec(\frac{\pi}{2} - x)$

know more about Trigonometry,

https://brainly.in/question/8632966

https://brainly.in/question/11371684

#SPJ2